K. Webb ESE 499
2
Introduction
We have seen how to design feedback control systems using the root locus
In this section of the course, weβll learn how to do the same using the open-loop frequency response
Objectives: Determine static error constants from the open-loop
frequency response Determine closed-loop stability from the open-loop
frequency response Use the open-loop frequency response for compensator
design to: Improve steady-state error Improve transient response
K. Webb ESE 499
4
Static Error Constants
For unity-feedback systems, open-loop transfer function gives static error constants Use static error constants to calculate steady-state
errorπΎπΎππ = lim
π π β0πΊπΊ π π
πΎπΎπ£π£ = limπ π β0
π π πΊπΊ π π
πΎπΎππ = limπ π β0
π π 2πΊπΊ π π
We can also determine static error constants from a systemβs open-loop Bode plot
K. Webb ESE 499
5
Static Error Constant β Type 0
For a type 0 system
πΎπΎππ = limπ π β0
πΊπΊ π π
At low frequency, i.e. below any open-loop poles or zeros
πΊπΊ π π β πΎπΎππ
Read πΎπΎππ directly from the open-loop Bode plot Low-frequency gain
πΊπΊ π π =100 π π + 30π π + 3 π π + 200
K. Webb ESE 499
6
Static Error Constant β Type 1
For a type 1 systemπΎπΎπ£π£ = lim
π π β0π π πΊπΊ π π
At low frequencies, i.e. below any other open-loop poles or zeros
πΊπΊ π π β πΎπΎπ£π£π π
and πΊπΊ ππππ β πΎπΎπ£π£ππ
A straight line with a slope of β20 ππππ/ππππππ Evaluating this low-frequency asymptote at ππ = 1
yields the velocity constant, πΎπΎπ£π£ On the Bode plot, extend the low-frequency asymptote
to ππ = 1 Gain of this line at ππ = 1 is πΎπΎπ£π£
K. Webb ESE 499
7
Static Error Constant β Type 1
πΊπΊ π π =85 π π + 0.1 π π + 50π π π π 2 + 10π π + 125
K. Webb ESE 499
8
Static Error Constant β Type 2
For a type 2 systemπΎπΎππ = lim
π π β0π π 2πΊπΊ π π
At low frequencies, i.e. below any other open-loop poles or zeros
πΊπΊ π π β πΎπΎπππ π 2
and πΊπΊ ππππ β πΎπΎππππ2
A straight line with a slope of β40 ππππ/ππππππ Evaluating this low-frequency asymptote at ππ = 1
yields the acceleration constant, πΎπΎππ On the Bode plot, extend the low-frequency asymptote
to ππ = 1 Gain of this line at ππ = 1 is πΎπΎππ
K. Webb ESE 499
9
Static Error Constant β Type 2
πΊπΊ π π =1600 π π + 0.1 π π + 5
π π 2 π π + 100
K. Webb ESE 499
11
Stability
Consider the following system
We already have a couple of tools for assessing stability as a function of loop gain, πΎπΎ Routh Hurwitz Root locus
Root locus: Stable for some values of πΎπΎ Unstable for others
K. Webb ESE 499
12
Stability
In this case gain is stable below some value
Other systems may be stable for gain abovesome value
Marginal stability point: Closed-loop poles on the
imaginary axis at Β±ππππ1 For gain πΎπΎ = πΎπΎ1
K. Webb ESE 499
13
Open-Loop Frequency Response & Stability
Marginal stability point occurs when closed-loop poles are on the imaginary axis Angle criterion satisfied at Β±ππππ1
πΎπΎπΊπΊ ππππ1 = 1 and β πΎπΎπΊπΊ ππππ1 = β180Β°
Note that β180Β° = 180Β°
πΎπΎπΊπΊ ππππ is the open-loop frequency response Marginal stability occurs when:
Open-loop gain is: πΎπΎπΊπΊ ππππ = 0 ππππ Open-loop phase is: β πΎπΎπΊπΊ ππππ = β180Β°
K. Webb ESE 499
14
Stability from Open-Loop Bode Plots
Here, stable for smaller gain values πΎπΎπΊπΊ ππππ < 0 ππππ whenβ πΎπΎπΊπΊ ππππ = β180Β°
Often, stable for larger gain values πΎπΎπΊπΊ ππππ > 0 ππππ whenβ πΎπΎπΊπΊ ππππ = β180Β°
Root locus provides this information Bode plot does not
Varying πΎπΎ simply shifts gain response up or down
K. Webb ESE 499
15
Open-Loop Frequency Response & Stability
Open-loop Bode plot can be used to assess stability But, we need to know if system is closed-loop stable for low gain
or high gain
Here, weβll assume open-loop-stable systems Closed-loop stable for low gain
Open-loop Bode plot can tell us: Is a system closed-loop stable? If so, how stable? I.e. how close to marginal stability
Two stability metrics: Gain margin Phase margin
K. Webb ESE 499
17
Crossover Frequencies
Two important frequencies when assessing stability:
Gain crossover frequency, ππππππ The frequency at
which the open-loop gain crosses 0 ππππ
Phase crossover frequency, πππΊπΊππ The frequency at
which the open-loop phase crosses β180Β°
K. Webb ESE 499
18
Gain Margin
An open-loop-stable system will be closed-loop stable as long as its gain is less than unity at the phase crossover frequency
Gain margin, GM The change in open-
loop gain at the phase crossover frequency required to make the closed-loop system unstable
K. Webb ESE 499
19
Phase Margin
An open-loop-stable system will be closed-loop stable as long as its phase has not fallen below β180Β° at the gain crossover frequency
Phase margin, PM The change in open-
loop phase at the gain crossover frequency required to make the closed-loop system unstable
K. Webb ESE 499
21
Phase Margin and Damping Ratio, ππ PM can be expressed as a function of damping ratio, ππ, as
ππππ = tanβ1 2ππ
β2ππ2+ 1+4ππ4
For ππππ β€ 65Β° or so, we can approximate:
ππππ β 100ππ or ππ β ππππ100
K. Webb ESE 499
23
bode.m
[mag,phase] = bode(sys,w)
sys: system model β state-space, transfer function, or other w: optional frequency vector β in rad/sec mag: system gain response vector phase: system phase response vector β in degrees
If no outputs are specified, bode response is automatically plotted β preferable to plot yourself
Frequency vector input is optional If not specified, MATLAB will generate automatically
May need to do: squeeze(mag) and squeeze(phase)to eliminate singleton dimensions of output matrices
K. Webb ESE 499
24
margin.m
[GM,PM,wgm,wpm] = margin(sys)
sys: system model β state-space, transfer function, or other GM: gain margin PM: phase margin β in degrees wgm: frequency at which GM is measured, the phase crossover
frequency β in rad/sec wpm: frequency at which PM is measured, the gain crossover
frequency
If no outputs are specified, a Bode plot with GM and PM indicated is automatically generated
K. Webb ESE 499
26
Frequency-Response Design
In a previous section of notes, we saw how we can use root-locus techniques to design compensators
Two primary objectives of compensation Improve steady-state error Proportional-integral (PI) compensation Lag compensation
Improve dynamic response Proportional-derivative (PD) compensation Lead compensation
Now, weβll learn to design compensators using a systemβs open-loop frequency response Weβll focus on lag and lead compensation
K. Webb ESE 499
28
Improving Steady-State Error
Consider the system above with a desired phase margin of ππππ β 50Β°
According to the Bode plot: ππ = β130Β° at ππππππ = 3.46 ππππππ/π π ππππ
Gain is πΎπΎππππ = β12.1 ππππat ππππππ
Set πΎπΎ = βπΎπΎππππ = 12.1ππππ = 4for desired phase margin
K. Webb ESE 499
29
Improving Steady-State Error
Can read the position constant directly from the Bode plot: πΎπΎππ = 14.8 ππππ β 5.5
Note that ππππ β50Β°, as desired
Gain margin is πΊπΊππ = 17.9 ππππ
K. Webb ESE 499
30
Improving Steady-State Error
Steady-state error to a constant reference is
πππ π π π =1
1 + πΎπΎππ= 0.154 β 15.4%
K. Webb ESE 499
31
Improving Steady-State Error
Letβs say we want to reduce steady-state error to πππ π π π <5%
Required position constant
πΎπΎππ >1
0.05 β 1 = 19
Increase gain by 4x Bode plot shows
desired position constant
But, phase margin has been degraded significantly
K. Webb ESE 499
32
Improving Steady-State Error
Step response shows that error goal has been met But, reduced phase margin results in significant overshoot
and ringing Error improvement came
at the cost of degraded phase margin
Would like to be able to improve steady-state error without affecting phase margin Integral compensation Lag compensation
K. Webb ESE 499
34
PI Compensation
Proportional-integral (PI) compensator:
π·π· π π =1πππ·π·
πππ·π·π π + 1π π
Low-frequency gain increase Infinite at DC System type increase
For ππ β« 1/πππ·π· Gain unaffected Phase affected little PM unaffected
Susceptible to integrator overflow Lag compensation is often
preferable
K. Webb ESE 499
36
Lag Compensation
Lag compensator
π·π· π π = πΌπΌπππ π + 1πΌπΌπππ π + 1
, πΌπΌ > 1
Objective: add a gain of πΌπΌ at low frequencies without affecting phase margin
Lower-frequency pole: π π = β1/πΌπΌππ Higher-frequency zero: π π = β1/ππ Pole/zero spacing determined by πΌπΌ For ππ βͺ 1/πΌπΌππ
Gain: ~20 log πΌπΌ ππππ Phase: ~0Β°
For ππ β« 1/ππ Gain: ~0 ππππ Phase: ~0Β°
K. Webb ESE 499
37
Lag Compensation vs. πΌπΌ
Gain increased at low frequency only Dependent on πΌπΌ DC gain: 20log πΌπΌ ππππ
Phase lag added between compensator pole and zero 0Β° β€ ππππππππ β€ 90Β° Dependent on πΌπΌ
Lag pole/zero well below crossover frequency Phase margin unaffected
π·π· π π = πΌπΌπππ π + 1πΌπΌπππ π + 1
K. Webb ESE 499
38
Lag Compensator Design Procedure
Lag compensator adds gain at low frequencies without affecting phase margin
Basic design procedure: Adjust gain to achieve the desired phase margin Add compensation, increasing low-frequency gain to
achieve desired error performance
Same as adjusting gain to place poles at the desired damping on the root locus, then adding compensation Root locus is not changed Here, the frequency response near the crossover frequency
is not changed
K. Webb ESE 499
39
Lag Compensator Design Procedure
1. Adjust gain, πΎπΎ, of the uncompensated system to provide the desired phase margin plus 5Β° β¦ 10Β° (to account for small phase lag added by compensator)
2. Use the open-loop Bode plot for the uncompensated system with the value of gain set in the previous step to determine the static error constant
3. Calculate πΆπΆ as the low-frequency gain increase required to provide the desired error performance
4. Set the upper corner frequency (the zero) to be one decade below the crossover frequency: 1/ππ = ππππππ/10 Minimizes the added phase lag at the crossover frequency
5. Calculate the lag pole: 1/πΌπΌππ6. Simulate and iterate, if necessary
K. Webb ESE 499
40
Lag Example β Step 1
Design a lag compensator for the above system to satisfy the following requirements πππ π π π < 2% for a step input %ππππ β 12%
First, determine the required phase margin to satisfy the overshoot requirement
ππ = βln ππππ
ππ2 + ln2 ππππ= 0.559
ππππ β 100ππ = 55.9Β° Add ~10Β° to account for compensator phase at ππππππ
ππππ = 65.9Β°
K. Webb ESE 499
41
Lag Example β Step 1
Plot the open-loop Bode plot of the uncompensated system for πΎπΎ = 1
Locate frequency where phase isβ180Β° + ππππ = β114.1Β° This is ππππππ, the desired
crossover frequency ππππππ = 2.5 ππππππ/π π ππππ
Gain at ππππππ is πΎπΎππππ πΎπΎππππ = β8.4 ππππ β 0.38
Increase the gain by 1/πΎπΎππππ πΎπΎ = 8.4 ππππ β 2.63
K. Webb ESE 499
42
Lag Example β Step 2
Gain has now been set to yield the desired phase margin of ππππ = 65.9Β°
Use the new open-loop bode plot to determine the static error constant
Position constant of the uncompensated system given by the DC gain:
πΎπΎππππ = 11.14 ππππ β 3.6
K. Webb ESE 499
43
Lag Example β Step 3
Calculate πΌπΌ to yield desired steady-state error improvement Steady-state error:
πππ π π π =1
1 + πΎπΎππ< 0.02
The required position constant:
πΎπΎππ >1πππ π π π
β 1 = 49 β πΎπΎππ = 50
Calculate πΌπΌ as the required position constant improvement
πΌπΌ =πΎπΎπππΎπΎππππ
= 13.9 β πΌπΌ = 14
K. Webb ESE 499
44
Lag Example β Steps 4 & 5
Place the compensator zero one decade below the crossover frequency, ππππππ = 2.5 ππππππ/π π ππππ
1/ππ = 0.25 ππππππ/π π ππππππ = 4 π π ππππ
The compensator pole:
1/πΌπΌππ = 0.2514
1/πΌπΌππ = 0.018 ππππππ/π π ππππ
Lag compensator transfer function
π·π· π π = πΌπΌπππ π + 1πΌπΌπππ π + 1
π·π· π π = 144π π + 156π π + 1
K. Webb ESE 499
45
Lag Example β Step 6
Bode plot of compensated system shows:
ππππ = 60.5Β°πΎπΎππ = 50.5
K. Webb ESE 499
46
Lag Example β Step 6
Lag compensator adds gain at low frequencies only
Phase near the crossover frequency is nearly unchanged
K. Webb ESE 499
47
Lag Example β Step 6
Steady-state error requirement has been satisfied
Overshoot spec has been met Though slow tail
makes overshoot assessment unclear
K. Webb ESE 499
48
Lag Compensator β Summary
π·π· π π = πΌπΌπππ π + 1πΌπΌπππ π + 1
Higher-frequency zero: π π = β1/ππ Place one decade below crossover frequency, ππππππ
Lower-frequency pole: π π = β1/πΌπΌππ πΌπΌ sets pole/zero spacing
DC gain: πΌπΌ β 20 log10 πΌπΌ ππππ
Compensator adds low-frequency gain Static error constant improvement Phase margin unchanged
K. Webb ESE 499
50
Improving Dynamic Response
Weβve already seen two types of compensators to improve dynamic response Proportional derivative (PD) compensation Lead compensation
Unlike with the lag compensator we just looked at, here, the objective is to alter the open-loop phase
Weβll look briefly at PD compensation, but will focus on lead compensation
K. Webb ESE 499
52
PD Compensation
Proportional-Derivative (PD) compensator:π·π· π π = πππ·π·π π + 1
Phase added near (and above) the crossover frequency Increased phase margin Stabilizing effect
Gain continues to rise at high frequencies Sensor noise is amplified Lead compensation is
usually preferable
K. Webb ESE 499
54
Lead Compensation
With lead compensation, we have three design parameters: Crossover frequency, ππππππ Determines closed-loop bandwidth, πππ΅π΅π΅π΅; risetime, π‘π‘ππ; peak time, π‘π‘ππ; and settling time, π‘π‘π π
Phase margin, PM Determines damping, ππ, and overshoot
Low-frequency gain Determines steady-state error performance
Weβll look at the design of lead compensators for two common scenarios, either Designing for steady-state error and phase margin, or Designing for closed-loop bandwidth and phase margin
K. Webb ESE 499
55
Lead Compensation
Lead compensator
π·π· π π =πππ π + 1π½π½πππ π + 1
, π½π½ < 1
Objectives: add phase lead near the crossover frequency and/or alter the crossover frequency
Lower-frequency zero: π π = β1/ππ Higher-frequency pole: π π = β1/π½π½ππ Zero/pole spacing determined by π½π½ For ππ βͺ 1/ππ
Gain: ~0 ππππ Phase: ~0Β°
For ππ β« 1/π½π½ππ Gain: ~20 log 1/π½π½ ππππ Phase: ~0Β°
K. Webb ESE 499
56
Lead Compensation vs. π½π½
π·π· π π =πππ π + 1π½π½πππ π + 1
, π½π½ < 1
π½π½ determines: Zero/pole spacing
Maximum compensator phase lead, ππππππππ
High-frequency compensator gain
K. Webb ESE 499
57
Lead Compensation β ππππππππ π½π½, zero/pole spacing, determines maximum phase lead
ππππππππ = sinβ11 β π½π½1 + π½π½
Can use a desired ππππππππ to determine π½π½
π½π½ =1 β sin ππππππππ1 + sin ππππππππ
ππππππππ occurs at ππππππππ
ππππππππ =1
ππ π½π½
ππ =1
ππππππππ π½π½
K. Webb ESE 499
58
Lead Compensation β Design Procedure
1. Determine loop gain, πΎπΎ, to satisfy either steady-state error requirements or bandwidth requirements:a) Set πΎπΎ to provide the required static error constant, orb) Set πΎπΎ to place the crossover frequency an octave below the desired
closed-loop bandwidth
2. Evaluate the phase margin of the uncompensated system, using the value of πΎπΎ just determined
3. If necessary, determine the required PM from ππ or overshoot specifications. Evaluate the PM of the uncompensated system and determine the required phase lead at the crossover frequency to achieve this PM. Add ~10Β° additional phase β this is ππππππππ
4. Calculate π½π½ from ππππππππ5. Set ππππππππ = ππππππ. Calculate ππ from ππππππππ and π½π½6. Simulate and iterate, if necessary
K. Webb ESE 499
59
Closed-Loop Bandwidth and Transient Response
Closed-loop bandwidth, πππ΅π΅π΅π΅, is one possible design criterion How is it related to transient response?
For a second-order system (or approximate second-order system): Closed-loop bandwidth and damping ratio and natural frequency, ππ and ππππ
πππ΅π΅π΅π΅ = ππππ 1 β 2ππ2 + 4ππ4 β 4ππ2 + 2
Closed-loop bandwidth and Β±1% settling time, π‘π‘π π
πππ΅π΅π΅π΅ β4.6π‘π‘π π ππ
1 β 2ππ2 + 4ππ4 β 4ππ2 + 2
Closed-loop bandwidth and peak time, π‘π‘ππ
πππ΅π΅π΅π΅ =4
π‘π‘ππ 1 β ππ21 β 2ππ2 + 4ππ4 β 4ππ2 + 2
K. Webb ESE 499
60
Double-Lead Compensation
A lead compensator can add, at most, 90Β° of phase lead
If more phase is required, use a double-lead compensator
π·π· π π =πππ π + 1π½π½πππ π + 1
2
For phase lead over ~60Β° β¦ 70Β°, 1/π½π½ must be very large, so typically use double-lead compensation
K. Webb ESE 499
61
Lead Compensation β Example 1
Consider the following system
Design a compensator to satisfy the following πππ π π π < 0.1 for a ramp input%ππππ < 15%
Here, weβll design a lead compensator to simultaneously adjust low-frequency gain and phase margin
K. Webb ESE 499
62
Lead Example 1 β Steps 1 & 2
The velocity constant for the uncompensated system isπΎπΎπ£π£ = lim
π π β0π π πΎπΎπΊπΊ π π
πΎπΎπ£π£ = limπ π β0
πΎπΎπ π + 1 = πΎπΎ
Steady-state error is
πππ π π π =1πΎπΎπ£π£
< 0.1
πΎπΎπ£π£ = πΎπΎ > 10 Adding a bit of margin
πΎπΎ = 12 Bode plot shows the resulting
phase margin is ππππ = 16.4Β°
K. Webb ESE 499
63
Lead Example 1 β Step 3
Approximate required phase margin for %ππππ < 15% Design for 13%
First calculate the required damping ratio
ππ = βln ππππ
ππ2 + ln2 ππππ= 0.545
Approximate corresponding PM, and add 10Β° correction factor
ππππ β 100ππ + 10Β° = 64.5Β°
Calculate the required phase lead
ππππππππ = 64.5Β° β 16.4Β° = 48Β°
K. Webb ESE 499
64
Lead Example 1 β Steps 4 & 5
Calculate π½π½ from ππππππππ
π½π½ =1 β sin ππππππππ1 + sin ππππππππ
= 0.147
Set ππππππππ = ππππππ, as determined from Bode plot, and calculate ππ
ππππππππ = ππππππ = 3.4 ππππππ/π π ππππ
ππ = 1ππππππππ π½π½
= 13.4 0.147
= 0.766
The resulting lead compensator transfer function is
πΎπΎπ·π· π π = πΎπΎπππ π + 1π½π½πππ π + 1
= 120.766π π + 10.113π π + 1
K. Webb ESE 499
65
Lead Example 1 β Step 6
πΎπΎπ·π· π π = 120.766π π + 10.113π π + 1
The lead compensator Bode plot
K. Webb ESE 499
66
Lead Example 1 β Step 6
Lead-compensated system: ππππ = 48.5Β° ππππππ = 7.2 ππππππ/π π ππππ
High-frequency compensator gain increased the crossover frequency Phase was added at the
previous crossover frequency PM is below target
Move lead zero/pole to higher frequencies Reduce the crossover
frequency increase Improve phase margin
K. Webb ESE 499
67
Lead Example 1 β Step 6
As predicted by the insufficient phase margin, overshoot exceeds the target %ππππ = 20.9% > 15%
Redesign compensator for higher ππππππππ Improve phase margin Reduce overshoot
K. Webb ESE 499
68
Lead Example 1 β Step 6
The steady-state error requirement has been satisfied πππ π π π = 0.08 < 0.1
Will not change with compensator redesign Low-frequency gain
will not be changed
K. Webb ESE 499
69
Lead Example 1 β Step 6
Iteration yields acceptable value for ππππππππ ππππππππ = 5.5 rad/sec Maintain same zero/pole spacing, π½π½, and, therefore, same ππππππππ
Recalculate zero/pole time constants:
ππ =1
ππππππππ π½π½=
15.5 0.147
= 0.4742
π½π½ππ = 0.147 β 0.4742 = 0.0697
The updated lead compensator transfer function:
π·π· π π = 120.4742π π + 10.0697π π + 1
K. Webb ESE 499
70
Lead Example 1 β Step 6
Crossover frequency has been reduced ππππππ = 5.58 ππππππ/π π ππππ
Phase margin is close to the target ππππ = 58.2Β°
Dip in phase is apparent, because ππππππππ is now placed at point of lower open-loop phase
K. Webb ESE 499
71
Lead Example 1 β Step 6
Overshoot requirement now satisfied
%ππππ = 14.7% < 15%
Low-frequency gain has not been changed, so error requirement is still satisfied
Design is complete
K. Webb ESE 499
72
Lead Compensation β Example 2
Again, consider the same system
Design a compensator to satisfy the following π‘π‘π π β 1.2 π π ππππ (Β±1%)%ππππ β 10%
Now, weβll design a lead compensator to simultaneously adjust closed-loop bandwidth and phase margin
K. Webb ESE 499
73
Lead Example 2 β Step 1
The required damping ratio for 10% overshoot is
ππ = β ln ππππππ2+ln2 ππππ
= 0.5912
Given the required damping ratio, calculate the required closed-loop bandwidth to yield the desired settling time
πππ΅π΅π΅π΅ = 4.6π‘π‘π π ππ
1 β 2ππ2 + 4ππ4 β 4ππ2 + 2
πππ΅π΅π΅π΅ = 7.52 ππππππ/π π ππππ
Weβll initially set the gain, πΎπΎ, to place the crossover frequency, ππππππ, one octave below the desired closed-loop bandwidth
ππππππ = πππ΅π΅π΅π΅/2 = 3.8 ππππππ/π π ππππ
K. Webb ESE 499
74
Lead Example 2 β Step 1
Plot the Bode plot for πΎπΎ = 1 Determine the loop gain at
the desired crossover frequency
πΎπΎππππ = β23.3 ππππ
Adjust πΎπΎ so that the loop gain at the desired crossover frequency is 0 ππππ
πΎπΎ =1πΎπΎππππ
= 23.3 ππππ = 14.7
K. Webb ESE 499
75
Lead Example 2 β Steps 2 & 3
Phase margin for the uncompensated system:
ππππππ = 14.9Β°
Required phase margin to satisfy overshoot requirement:
ππππ β 100ππ = 59.1Β°
Add 10Β° to account for crossover frequency increase
ππππ = 69.1Β°
Required phase lead from the compensator
ππππππππ = ππππ β ππππππ = 54.2Β°
Generate a Bode plot using the gain value just determined
K. Webb ESE 499
76
Lead Example 2 β Steps 4 & 5
Calculate zero/pole spacing, π½π½, from required phase lead, ππππππππ
π½π½ =1 β sin ππππππππ1 + sin ππππππππ
= 0.1040
Calculate zero and pole time constants
ππ = 1ππππππππ π½π½
= 0.8228 π π ππππ
π½π½ππ = 0.0855 π π ππππ The resulting lead compensator transfer
function:
πΎπΎπ·π· π π = πΎπΎπππ π + 1π½π½πππ π + 1
πΎπΎπ·π· π π = 14.70.8228π π + 10.0855π π + 1
K. Webb ESE 499
77
Lead Example 2 β Step 6
Bode plot of the compensated system ππππ = 49.8Β° Substantially below
target
Crossover frequency is well above the desired valueππππππ = 9.44 ππππππ/π π ππππ
Iteration will likely be required
K. Webb ESE 499
78
Lead Example 2 β Step 6
Overshoot exceeds the specified limit %ππππ = 19.1Β° > 10%
Settling time is faster than required π‘π‘π π = 0.98 π π ππππ < 1.2 π π ππππ
Iteration is required Start by reducing the
target ππππππ
K. Webb ESE 499
79
Lead Example 2 β Step 6
Must redesign the compensator to meet specifications Must increase PM to reduce overshoot Can afford to reduce crossover, ππππππ, to improve PM
Try various combinations of the following Reduce crossover frequency, ππππππ Increase compensator zero/pole frequencies, ππππππππ Increase added phase lead, ππππππππ, by reducing π½π½
Iteration shows acceptable results for: ππππππ = 2.4 ππππππ/π π ππππ ππππππππ = 3.4 ππππππ/π π ππππ ππππππππ = 52Β°
K. Webb ESE 499
80
Lead Example 2 β Step 6
Redesigned lead compensator:
πΎπΎπ·π· π π = 6.270.8542π π + 10.1013π π + 1
Phase margin:ππππ = 62Β°
Crossover frequency:ππππππ = 4.84 ππππππ/π π ππππ
K. Webb ESE 499
81
Lead Example 2 β Step 6
Dynamic response requirements are now satisfied
Overshoot:%ππππ = 8%
Settling time:π‘π‘π π = 1.09 π π ππππ
K. Webb ESE 499
82
Lead Compensation β Example 2
Lead compensator adds gain at higher frequencies Increased crossover
frequency Faster response time
Phase added near the crossover frequency Improved phase
margin Reduced overshoot
K. Webb ESE 499
83
Lead Compensation β Example 2
Step response improvements: Faster settling time Faster risetime Significantly less
overshoot and ringing
K. Webb ESE 499
84
Lead-Lag Compensation
If performance specifications require adjustment of: Bandwidth Phase margin Steady-state error
Lead-lag compensation may be used
π·π· π π = πΌπΌπππππππππ π + 1πΌπΌπππππππππ π + 1
πππππππππππ π + 1π½π½πππππππππππ π + 1
Many possible design procedures β one possibility:1. Design lag compensation to satisfy steady-state error and
phase margin2. Add lead compensation to increase bandwidth, while
maintaining phase margin